Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{8z^2 - 24z}{-7z + 70} \div \dfrac{z^2 - 12z + 27}{z - 10} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{8z^2 - 24z}{-7z + 70} \times \dfrac{z - 10}{z^2 - 12z + 27} $ First factor the quadratic. $q = \dfrac{8z^2 - 24z}{-7z + 70} \times \dfrac{z - 10}{(z - 3)(z - 9)} $ Then factor out any other terms. $q = \dfrac{8z(z - 3)}{-7(z - 10)} \times \dfrac{z - 10}{(z - 3)(z - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 8z(z - 3) \times (z - 10) } { -7(z - 10) \times (z - 3)(z - 9) } $ $q = \dfrac{ 8z(z - 3)(z - 10)}{ -7(z - 10)(z - 3)(z - 9)} $ Notice that $(z - 10)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 8z\cancel{(z - 3)}(z - 10)}{ -7(z - 10)\cancel{(z - 3)}(z - 9)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $q = \dfrac{ 8z\cancel{(z - 3)}\cancel{(z - 10)}}{ -7\cancel{(z - 10)}\cancel{(z - 3)}(z - 9)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $q = \dfrac{8z}{-7(z - 9)} $ $q = \dfrac{-8z}{7(z - 9)} ; \space z \neq 3 ; \space z \neq 10 $